Solve the equation for x. show all work. ?
y = 8cos(3x) y/8 = cos(3x) arccos(y/8) = 3x x = arccos(y/8)/3
Solve the following equation. Show all work.?
3x/5 = +7-13 = -6 so, 3x = -6*5 so, x = -6*5/3 giving x = -10
Solve the system of equations algebraically and show your work.?
The objective in solving two equations with two unknowns (x and y in this case) is to first solve for one of the two unknowns and then solve for the other. The way to do this is to add or subtract the two equations in some way so that you are left with only x or only one y equaling some number. Once you have solved the first unknown, you can then calculate the other unknown simply by placing the newly calculated value into either one of the two equations. Let's try this ... First let's eliminate the x's in the two equations. The way to do this is by subtraction. However, the first equation has "3x" and the second has only one "x". In order to subtract the two equations so that the x's cancel, we need to multiple the second equation by 3 (so that x becomes 3x). You may have learned in school that when you multiple one side of an equation by a number you also have to multiple the other side. So when you multiple the second equation by 3, all terms are multiplied by 3. That is, 3 * (x + 4y = -8) becomes 3x + 12y = -24. Now we can subtract the two equations because the x's will cancel and we are left with only y = some number that we can solve: (3x + 2y = 6) - (3x + 12y = -24) = -10y = 30. Now we can solve this equation for y: -10y = 30 y = -3 Now, all we need to do is calculate x. Substituting this value of y into either one of the equations will calculate the x value. Let's use the first equation: 3x + 2y = 6. 3x + 2(-3) = 6 Solving for x: 3x = 12 x = 4 So our x and y value are: x = 4 and y = -3. To check our work, it is good practice to enter the calculated values of x and y into the other equation to make sure that the equation equality holds. Let's try that: x + 4y = -8 4 + 4(-3) = -8 4 + (-12) = -8 -8 = -8 Great! So we have confirmed that our calculated values of x and y work in both equations. I hope this was helpful to you.
Please help me solve literal equation with work shown.?
x+y=10 for y (I believe anwer is y=10-x) Yea, in order to show ur work subtract the X on both sides 4x=y-16 for x uh....there'll be a Y in ur answer but u just divide both sides by 4, so u get x= 1/4Y - 4 M=1/2DC for C divide both sides by D, then u got MD = 1/2C, then multiby both sides by 2: 2MD= C p=4s for s( I believe answer is s=4P?) Divide both sides by 4 cuz that's the opposite of multiplication, so p/4 =s 7y=9x=21 for y if u just want y, ignore the 9x. 7y=21, divide both sides by 7, 21/7 = 3 e=mc(2) for m (this is mc squared) so u got e=mc times mc, first divide both by mc and them the c so u got e/mcc OK, basically all these probs u gave r the same, just get the letter by itself by doing the opposte. if it's being multiplied by a 3, divide by a 3. etc. u can do the rest, don't worry! 2a+3b=c for a 2y=3x+14 for x x+y=2 for y x-y=5 for y 2a+3b=c for b v=1wh for h 4x-3y=12 for y